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Convergence and Divergence of Sequences

We will now look at two very important terms when it comes to categorizing sequences.

Definition: A sequence $(a_n)$ is said to be convergent to the real number $L$ and we write $\lim_{n \to \infty} a_n = L$ if $\forall \epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$. If $\lim_{n \to \infty} a_n$ does not exist (that is the limit is infinity, negative infinity, or just doesn't converge in general) then we say that the sequence $(a_n)$ is divergent.

From this definition of convergence, we immediately have the following theorem of equivalence statements.

Theorem 1: Let $(a_n)$ be a convergent sequence. Then the following statements are equivalent:1. The sequence $(a_n)$ converges to the real number $L$.2. For every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$.3. For every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ the terms satisfy $L - \epsilon < a_n < L + \epsilon$.4. For every $\epsilon > 0$, for the $\epsilon$-neighbourhood $V_{\epsilon}(L)$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $x_n \in V_{\epsilon} (L)$.

Proof:$1 \Leftrightarrow 2$ Suppose that the sequence $(a_n )$ converges to the real number $L$. Then $\lim_{n \to \infty} a_n = L$ and by the definition, for every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - \mid < \epsilon$.

$2 \Leftrightarrow 3$ Suppose that for every $\epsilon > 0$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$. Taking the last inequality, by an earlier theorem we note that $\mid a_n - L \mid < \epsilon$ if and only if:

$3 \Leftrightarrow 4$ If for every $\epsilon > 0$, for the $\epsilon$-neighbourhood $V_{\epsilon}(L)$ there exists a natural number $N \in \mathbb{N}$ such that if $n ≥ N$ then $x_n \in V_{\epsilon} (L)$, then $x_n \in V_{\epsilon} (L)$ for all $n ≥ N$. $\blacksquare$

The following theorem tells us that the sequence $\displaystyle{\{ \frac{1}{n} \right \}}_{n=1}^{\infty}$ converges to $0$.

We want $\frac{1}{n} < \epsilon$, i.e., $\frac{1}{\epsilon} < n$. So choose $N > \frac{1}{\epsilon}$. Then if $n \geq N$ we have that $n > \frac{1}{\epsilon}$, and so $\frac{1}{n} < \epsilon$, so from above: